Parametric Arc Length

**apples12** · October 2nd 2009, 09:55 AM

Alright I'm pretty lost here. I can usually deal with these types of problems fine, but with the given equations I don't seem to be getting anywhere.

Given two parametric functions:

x(t)=5e−0

5tsin(2t)
y(t)=5e−0

5tcos(2t)

0

t

6.

Find the integral.
Find the arc length, 0

t

6.
Find the arc length, [0

).

Getting the derivative of each term and squaring it yields some nasty numbers and I'm running out of submissions.

The equation to go with that I've been using is:

**skeeter** · October 2nd 2009, 10:21 AM

Originally Posted by apples12

Alright I'm pretty lost here. I can usually deal with these types of problems fine, but with the given equations I don't seem to be getting anywhere.

Given two parametric functions:

x(t)=5e−05tsin(2t) are the bold expressions exponents?
y(t)=5e−05tcos(2t)

0

t

6.

Find the integral.
Find the arc length, 0

t

6.
Find the arc length, [0

).

Getting the derivative of each term and squaring it yields some nasty numbers and I'm running out of submissions.

The equation to go with that I've been using is:

...

**apples12** · October 2nd 2009, 10:35 AM

sorry my formatting got messed up in the posting

x(t)=sin(2t)5e^(-0.5t)
y(t)=cos(2t)5e^(-0.5t)

**Amer** · October 2nd 2009, 12:49 PM

$x(t) = 5 e^{t/2} \sin 2t$

$y(t) = 5 e^{t/2} \cos 2t$

$x' = 5(1/2 e^{t/2} \sin 2t + 2e^{t/2} \cos 2t)$

$x' = 5e^{t/2} (1/2 \sin 2t + 2\cos 2t )$

in the same way

$y' = 5e^{t/2} (1/2 \cos 2t - 2\sin 2t )$

ok

$x'^2 + y'^2 = 25e^t ( 1/4 \sin ^2 2t + 2\sin 2t \cos 2t + 4 \cos ^2 2t +$ $1/4 \cos ^2 2t - 2 \sin 2t \cos 2t + 4\sin ^2 2t )$

$x'^2 + y'^2 = 25e^t (1/4 + 4 )$ and this is easy to integrate

**apples12** · October 2nd 2009, 01:35 PM

this is wrong, the exponents are negative

**skeeter** · October 2nd 2009, 04:27 PM

Originally Posted by apples12

this is wrong, the exponents are negative

so? ... take Amer's post and fix the signs.

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